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Matrices and Determinants: Mathematics | JEE Main

Q: Does unique solution depend on ?

No, for non-homogeneous systems, alone guarantees a unique solution, regardless of the value of .

Consider the system of linear equation

x+y+z=4 \mu, x+2 y+2 \lambda z=10 \mu, x+3 y+4 \lambda^{2} z=\mu^{2}+15

, where

\lambda, \quad \mu \in \mathrm{R}

. Which one of the following statements is NOT correct?

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

AThe system has unique solution if

\lambda \neq \frac{1}{2}

and

\mu \neq 1,15

BThe system is inconsistent if

\lambda=\frac{1}{2}

and

\mu \neq 1

CThe system has infinite number of solutions if

\lambda=\frac{1}{2}

and

\mu=15

DThe system is consistent if

\lambda \neq \frac{1}{2}

Step-by-step solutionView

Correct answer

The incorrect statement is that the system is inconsistent if $\lambda = \frac{1}{2}$ and $\mu \neq 1$ , because the system is actually inconsistent for $\lambda = \frac{1}{2}$ whenever $\mu \neq 1$ AND $\mu \neq 15$ (at $\mu = 15$ , it has infinitely many solutions).

Option analysis

Why each option works or fails

A · The system has unique solution if

\lambda \neq \frac{1}{2}

and

\mu \neq 1,15

Believing this statement is false because $\mu$ is restricted unnecessarily, forgetting that any sub-condition where $\Delta \neq 0$ still guarantees a unique solution regardless of $\mu$ . When $\Delta = (2\lambda - 1)^2 \neq 0$ , the coefficient matrix is invertible, which guarantees a unique solution for all values of $\mu$ (including the specific subset where $\mu \neq 1, 15$ ).

B · The system is inconsistent if

\lambda=\frac{1}{2}

and

\mu \neq 1

Assuming that $\mu \neq 1$ is sufficient to make the system inconsistent when $\lambda = \frac{1}{2}$ , overlooking the root $\mu = 15$ which also makes the system consistent with infinitely many solutions. Row reduction shows that when $\lambda = \frac{1}{2}$ , the last equation becomes $0 = (\mu - 1)(\mu - 15)$ . Thus, the system remains consistent at $\mu = 15$ , making the claim that it is inconsistent for all $\mu \neq 1$ false.

C · The system has infinite number of solutions if

\lambda=\frac{1}{2}

and

\mu=15

Confusing infinite solutions with inconsistency when determinant is zero. Substitute $\lambda = \frac{1}{2}$ and $\mu = 15$ into the augmented matrix: the rank of the coefficient matrix and the augmented matrix both equal 2, confirming infinitely many solutions.

D · The system is consistent if

\lambda \neq \frac{1}{2}

Believing consistency requires conditions on $\mu$ even when the determinant of the coefficient matrix is non-zero. Whenever $\lambda \neq \frac{1}{2}$ , $\det(A) \neq 0$ , ensuring the system has a unique solution and is therefore consistent for every real $\mu$ .

Reviewed route

Solution

StepWorking

01given

System of equations: 1) $x + y + z = 4\mu$ 2) $x + 2y + 2\lambda z = 10\mu$ 3) $x + 3y + 4\lambda^2 z = \mu^2 + 15$

02goal

Identify which of the four given statements regarding the consistency/solutions of the system is NOT correct.

03approach

Compute the coefficient determinant $\Delta$ . Find conditions on $\lambda$ for $\Delta \neq 0$ (unique solution) and $\Delta = 0$ . Then evaluate $\Delta_x, \Delta_y, \Delta_z$ at $\lambda = 1/2$ to find conditions on $\mu$ for infinite solutions versus no solution (inconsistency).

04execute

Calculate the coefficient determinant: $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 2\lambda \\ 1 & 3 & 4\lambda^2 \end{vmatrix}$ Perform row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$ : $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 0 & 1 & 2\lambda - 1 \\ 0 & 2 & 4\lambda^2 - 1 \end{vmatrix} = 1 \cdot [(4\lambda^2 - 1) - 2(2\lambda - 1)] = (2\lambda - 1)(2\lambda + 1 - 2) = (2\lambda - 1)^2$ Thus, $\Delta \neq 0 \iff \lambda \neq \frac{1}{2}$ . When $\lambda \neq \frac{1}{2}$ , the system has a unique solution for ALL $\mu \in \mathbb{R}$ .

05execute

For $\lambda = \frac{1}{2}$ , $\Delta = 0$ . Evaluate $\Delta_z$ (or $\Delta_x$ ): $\Delta_z = \begin{vmatrix} 1 & 1 & 4\mu \\ 1 & 2 & 10\mu \\ 1 & 3 & \mu^2 + 15 \end{vmatrix}$ Using $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$ : $\Delta_z = \begin{vmatrix} 1 & 1 & 4\mu \\ 0 & 1 & 6\mu \\ 0 & 2 & \mu^2 - 4\mu + 15 \end{vmatrix} = (\mu^2 - 4\mu + 15) - 2(6\mu) = \mu^2 - 16\mu + 15 = (\mu - 1)(\mu - 15)$ - For infinite solutions: $\Delta = \Delta_x = \Delta_y = \Delta_z = 0 \implies \lambda = \frac{1}{2}$ and $\mu = 1$ or $\mu = 15$ . - For inconsistent system: $\lambda = \frac{1}{2}$ and $\mu \neq 1, 15$ . Hence, Statement (1) claims 'The system is inconsistent if $\lambda = 1/2$ and $\mu \neq 1$ ', which is FALSE because when $\mu = 15$ , the system is consistent (has infinitely many solutions).

✓verify

Check statement (0): While it adds unnecessary conditions on $\mu$ , a system with $\lambda \neq 1/2$ and $\mu \neq 1, 15$ does indeed have a unique solution. However, statement (1) asserts inconsistency for all $\mu \neq 1$ , which directly fails at $\mu = 15$ . Thus (1) is definitively NOT correct.

Hints that build this answer step by step

What is the determinant of the coefficient matrix $A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 2\lambda \\ 1 & 3 & 4\lambda^2 \end{pmatrix}$ ?
$\det(A) = (2\lambda - 1)^2$
Applying row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - 2R_2 + R_1$ on the augmented matrix $[A | B]$ with $\lambda = \frac{1}{2}$ , what is the resulting third row?
$[0 \quad 0 \quad 0 \mid \mu^2 - 16\mu + 15]$
For $\lambda = \frac{1}{2}$ , for which values of $\mu$ is the system consistent (infinitely many solutions)?
$\mu = 1$ and $\mu = 15$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Does unique solution depend on

\mu

?

No, for non-homogeneous systems, $\Delta \neq 0$ alone guarantees a unique solution, regardless of the value of $\mu$ .

Answer

The incorrect statement is that the system is inconsistent if $\lambda = \frac{1}{2}$ and $\mu \neq 1$ , because the system is actually inconsistent for $\lambda = \frac{1}{2}$ whenever $\mu \neq 1$ AND $\mu \neq 15$ (at $\mu = 15$ , it has infinitely many solutions).

Why each option works or fails

A: The system has unique solution if $\lambda \neq \frac{1}{2}$ and $\mu \neq 1,15$ - Believing this statement is false because $\mu$ is restricted unnecessarily, forgetting that any sub-condition where $\Delta \neq 0$ still guarantees a unique solution regardless of $\mu$ . When $\Delta = (2\lambda - 1)^2 \neq 0$ , the coefficient matrix is invertible, which guarantees a unique solution for all values of $\mu$ (including the specific subset where $\mu \neq 1, 15$ ).
B · correct: The system is inconsistent if $\lambda=\frac{1}{2}$ and $\mu \neq 1$ - Assuming that $\mu \neq 1$ is sufficient to make the system inconsistent when $\lambda = \frac{1}{2}$ , overlooking the root $\mu = 15$ which also makes the system consistent with infinitely many solutions. Row reduction shows that when $\lambda = \frac{1}{2}$ , the last equation becomes $0 = (\mu - 1)(\mu - 15)$ . Thus, the system remains consistent at $\mu = 15$ , making the claim that it is inconsistent for all $\mu \neq 1$ false.
C: The system has infinite number of solutions if $\lambda=\frac{1}{2}$ and $\mu=15$ - Confusing infinite solutions with inconsistency when determinant is zero. Substitute $\lambda = \frac{1}{2}$ and $\mu = 15$ into the augmented matrix: the rank of the coefficient matrix and the augmented matrix both equal 2, confirming infinitely many solutions.
D: The system is consistent if $\lambda \neq \frac{1}{2}$ - Believing consistency requires conditions on $\mu$ even when the determinant of the coefficient matrix is non-zero. Whenever $\lambda \neq \frac{1}{2}$ , $\det(A) \neq 0$ , ensuring the system has a unique solution and is therefore consistent for every real $\mu$ .

Step-by-step solution

given: System of equations: 1) $x + y + z = 4\mu$ 2) $x + 2y + 2\lambda z = 10\mu$ 3) $x + 3y + 4\lambda^2 z = \mu^2 + 15$
goal: Identify which of the four given statements regarding the consistency/solutions of the system is NOT correct.
approach: Compute the coefficient determinant $\Delta$ . Find conditions on $\lambda$ for $\Delta \neq 0$ (unique solution) and $\Delta = 0$ . Then evaluate $\mu$ at $\lambda = 1/2$ to find conditions on $\mu$ for infinite solutions versus no solution (inconsistency).
execute: Calculate the coefficient determinant: $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 2\lambda \\ 1 & 3 & 4\lambda^2 \end{vmatrix}$ Perform row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$ : $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 0 & 1 & 2\lambda - 1 \\ 0 & 2 & 4\lambda^2 - 1 \end{vmatrix} = 1 \cdot [(4\lambda^2 - 1) - 2(2\lambda - 1)] = (2\lambda - 1)(2\lambda + 1 - 2) = (2\lambda - 1)^2$ Thus, $\Delta \neq 0 \iff \lambda \neq \frac{1}{2}$ . When $\lambda \neq \frac{1}{2}$ , the system has a unique solution for ALL $\mu \in \mathbb{R}$ .
execute: For $\lambda = \frac{1}{2}$ , $\Delta = 0$ . Evaluate $\Delta_z$ (or $\Delta_x$ ): $\Delta_z = \begin{vmatrix} 1 & 1 & 4\mu \\ 1 & 2 & 10\mu \\ 1 & 3 & \mu^2 + 15 \end{vmatrix}$ Using $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$ : $\Delta_z = \begin{vmatrix} 1 & 1 & 4\mu \\ 0 & 1 & 6\mu \\ 0 & 2 & \mu^2 - 4\mu + 15 \end{vmatrix} = (\mu^2 - 4\mu + 15) - 2(6\mu) = \mu^2 - 16\mu + 15 = (\mu - 1)(\mu - 15)$ - For infinite solutions: $\Delta = \Delta_x = \Delta_y = \Delta_z = 0 \implies \lambda = \frac{1}{2}$ and $\mu = 1$ or $\mu \in \mathbb{R}$ . - For inconsistent system: $\lambda = \frac{1}{2}$ and $\mu \neq 1, 15$ . Hence, Statement (1) claims 'The system is inconsistent if $\lambda = 1/2$ and $\mu \neq 1$ ', which is FALSE because when $\mu = 15$ , the system is consistent (has infinitely many solutions).
verify: Check statement (0): While it adds unnecessary conditions on $\mu$ , a system with $\lambda \neq 1/2$ and $\mu \neq 1, 15$ does indeed have a unique solution. However, statement (1) asserts inconsistency for all $\mu \neq 1$ , which directly fails at $\mu = 15$ . Thus (1) is definitively NOT correct.

Matrices and Determinants: Mathematics | JEE Main

What feels right?

The incorrect statement is that the system is inconsistent if $\lambda = \frac{1}{2}$ and $\mu \neq 1$ , because the system is actually inconsistent for $\lambda = \frac{1}{2}$ whenever $\mu \neq 1$ AND $\mu \neq 15$ (at $\mu = 15$ , it has infinitely many solutions).

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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